NCERT Solutions for Class 8 Chapter 11 Mensuration -Free PDF Download

Free PDF download of NCERT Solutions Maths Class 8 Solutions Chapter 11 – Mensuration solved by Expert Maths Teachers on CoolGyan.Org. All Chapter 11 – Mensuration Questions with Solutions for NCERT to help you to revise complete Syllabus and Score More marks.
Maths Revision Notes for Class 8

Chapter Name	Mensuration
Chapter	Chapter 11
Exercise	Exercise 11.3
Class	Class 8
Subject	Maths NCERT Solutions
Board	CBSE
TEXTBOOK	CBSE NCERT
Category	NCERT Solutions

NCERT SOLVED

1. There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?

Sol. (a) Length of cuboidal box = 60 cm

Breadth of cuboidal box = 40 cm

Height of cuboidal box = 50 cm

Total surface area of cuboidal box = $2 (l b + b h + h l)$

= 2 (60 40 + 40 50 + 50 60)

= 2 (2400 + 2000 + 3000)

= 2 7400

= 14800

(b) Length of the cube is 50 cm

Total surface area of cuboidal box = $6 (s i d e)^{2}$

=6 ( 50)²

= 6 (2500)

= 15000

Thus, the cuboidal box (a) requires the lesser amount of materal.

2. A suitcase with measures 80 cm 48 cm 24 cm is to be covered with a tarpaulin cloth. How many meters of tarpaulin of width 96 cm is required to cover 100 such suitcases?

Sol. Given: Length of suitcase box = 80 cm, Breadth of suitcase box = 48 cm

And Height of cuboidal box = 24 cm

Total surface area of suitcase box= $2 (l b + b h + h l)$

= 2 (80 48 + 48 24 + 24 80)

= 2 (3840 + 1152 + 1920)

= 2 6912 = 13824

Area of Tarpaulin cloth = Surface area of suitcase

= 13824

= 144 cm

Required tarpaulin for 100 suitcases = (144 100) cm

= 14400 cm

= 144 m [ 1cm = $\frac{1}{100}$ m]

Thus, 144 m tarpaulin cloth required to cover 100 suitcases.

3. Find the side of a cube whose surface area id 600.

Sol. Here Surface area of cube = 600 cm²
  = 600 cm²
  = 100 cm²
$l = \sqrt{1} 00$ cm
  = 10 cm
Hence the side of cube is 10 cm

4. Rukshar painted the outside of the cabinet of measure 1 m 2 m 1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet?

Sol. Length of cabinet = 2 m

Breadth of cabinet = 1 m

Height of cabinet = 1.5 m

Surface area of cabinet = (Area of Base of cabinet (Cuboid) + Area of four walls)

= $l b + 2 (l + b) h$
={2 1 + 2 (1 + 2) 1.5 }
= 2 + 2 (3) 1.5
=2+6 (1.5)
= (2 + 9.0)
= 11

Hence required surface area of cabinet is 11.

5. Daniel is paining the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10 m and 7 m respectively. From each can of paint 100 of area is painted. How many cans of paint will she need to paint the room?

Sol. Length of wall  = 15 m
Breadth of wall  = 10 m
Height of wall  = 7 m

Total Surface area of classroom=(Area of Base of ceiling (Cuboid) + Area of four walls)

= $l b + 2 (l + b) h$
= (15 10 + 2 (10 +15) (7))
= (150 + 2 (25) (7))
= (150 + 350)
= 500

Area of one can is 100 m²

Now Required number of cans = = 5 cans

Hence 5 cans are required to paint the room.

6. Describe how the two figures below are alike and how they are different. Which box has larger lateral surface area?

Sol.Diameter of cylinder = 7 cm
Radius of cylinder = cm
Height of cylinder = 7 cm
Lateral surface area of cylinder =
=
= 154 cm²

Now lateral surface area of cube = $4 (S i d e)^{2} = 4 (7)^{2}$
= (4 49)
= 196
Hence the cube has larger lateral surface area.

7. A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How much sheet of metal is required?

Sol. Radius of cylindrical tank = 7 m
Height of cylindrical tank = 3 m

Total surface area of cylindrical tank = (Curved surface area + Area of upper end (circle)+ Area of Lower (circle) end)

= $(2 π r h + π r^{2} + π r^{2})$

= $(2 π r h + 2 π r^{2})$

= $2 π r (h + r)$

=
= 44 10
= 440

Hence 440 metal sheet is required.

8. The lateral surface area of a hollow cylinder is 4224. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet?

Sol. Lateral surface area of hollow cylinder = 4224

Height of hollow cylinder = 33 cm

Curved surface area of hollow cylinder =

4224 =

= cm

Now Length of rectangular sheet =

= 128 cm

Perimeter of rectangular sheet =

= 2 (128 + 33)
= 2 x 161
= 322 cm

Hence perimeter of rectangular sheet is 322 cm.

9. A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm and length 1 m.

Sol. Diameter of road roller = 84 cm
Radius of road roller
= 42 cm

Length of road roller = 1 m = 100 cm

Curved surface area of road roller =
=
= 26400 cm²

Area covered by road roller in 750 revolutions = 26400 750
= 1,98,00,000
= 1980 m² [ 1 = 10,000]
Thus, the area of the road is 1980.

10. A company packages its milk powder in cylindrical container whose base has a diameter of 14 cm and height 20 cm. Company places a label around the surface of the container (as shown in figure). If the label is placed 2 cm from top and bottom, what is the area of the label?

Sol.Diameter of cylindrical container = 14 cm

Radius of cylindrical container = 7 cm

Height of cylindrical container = 20 cm

Height of the label = (20 – 2 – 2)
= 16 cm

Curved surface area of label =
= $2 \times \frac{22}{7} \times 7 \times 16$
= 704 cm²

Hence the area of the label of 704 cm².

NCERT Solutions for Class 8 Maths Chapter 11 (Ex 11.3) Mensuration

NCERT Solutions for Class 8 Chapter 11 Mensuration -Free PDF Download

NCERT SOLVED